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For proton and alpha-particle,de-Broglie...

For proton and `alpha`-particle,de-Broglie wavelength is same .For them……..will be equal .

A

velocity

B

energy

C

frequency

D

momentum

Text Solution

Verified by Experts

The correct Answer is:
D
`lambda_(p)=lambda_(alpha)implies(h)/(p_(p))=(h)/(p_(alpha))`
`therefore p_("proton")=p_(alpha)`
Momentum of proton and `alpha`-particle will be equal.
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